## Poll

 I love math! 19 votes (46.34%) Math is great. 14 votes (34.14%) My religion is mathology. 6 votes (14.63%) Women didn't speak to me until I was 30. 2 votes (4.87%) Total eclipse reminder -- 04/08/2024 12 votes (29.26%) I steal cutlery from restaurants. 3 votes (7.31%) I should just say what's on my mind. 6 votes (14.63%) Who makes up these awful names for pandas? 5 votes (12.19%) I like to touch my face. 11 votes (26.82%) Pork chops and apple sauce. 9 votes (21.95%)

41 members have voted

Wizard
Joined: Oct 14, 2009
• Posts: 24887
Thanks for this post from:
November 23rd, 2022 at 3:56:02 PM permalink

I show player 1 has an EV of \$0.10 and player 2 of \$0.15. Thus, player 2 is better off.

I might make a another version of this puzzle where collusion is allowed, but let's wait for this one to be closed first.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Gialmere
Joined: Nov 26, 2018
• Posts: 2472
November 23rd, 2022 at 5:44:57 PM permalink
Quote: charliepatrick

If a Red card is picked, #1 passes and #2 plays.
If a Black card is picked, #1 plays and #2 (perforce) passes.

As Red and Black cards are picked with equal probability, on average, #1 and #2 play the same number of games. Therefore we only need to look at which, on average, game gives the best return. For simplicity it's easier to sum up the five possible outcomes and see the total profit.

When #1 plays there are 3 Red and 2 Black, thus the profit is (3*\$1 - 2*\$1) = \$1.
When #2 plays there are 2 Red and 3 Black, however the cost to play is 50c so the profit when winning is \$1.50. Thus the profit is (2*\$1.50 - 3*50c) = \$1.50.

Hence #2 is the better off.

Quote: gordonm888

Player#1 can play with an EV = \$1 if he only play on when he picks up a black card, or an EV of zero if he plays no matter which color he picks up.

Player#2 can play only when Player#1 passes, but then he can make \$1.50 when Players#1 picked up a red card and more if Player picks up a black card but passes anyway.

However, Player #2 will never make any money unless palyer#1 passes.

So Player#1 charges Player#2 a fee in order to pass no matter what color card he sees. The fee might reasonably be set at 60% of what player#2 expects to make. Such an arrangement would be a win-win for both players.

So player#1 profits the most.

Quote: ThatDonGuy

This assumes that the \$2 that a player can win is the entire amount, as opposed to \$2 plus the amount bet.

If P1 sees a red card, and:
(a) plays, P1 has a 2/5 chance of winning \$2, for an ER of 4/5, and a EV of 4/5 - 1 = -1/5
(b) passes, P2 has a 2/5 chance of winning \$2, for an ER of 4/5, and a EV of 4/5 - 1/2 = 3/10

If P1 sees a black card, and:
(a) plays, P1 has a 3/5 chance of winning \$2, for an ER of 6/5, and a EV of 6/5 - 1 = 1/5
(b) passes, P2 has a 3/5 chance of winning \$2, for an ER of 6/5, and a EV of 6/5 - 1/2 = 7/10

The first card can equally likely be red or black. If it is red, P1's best play is to pass; if it is black, P1's best play is to play.
The EV for each hand is 1/2 x 0 + 1/2 x 1/5 = 1/10 for P1, and 1/2 x 3/10 + 1/2 x 0 = 3/20 for P2.
Player 2 will make more in the long run.

Quote: Wizard

I show player 1 has an EV of \$0.10 and player 2 of \$0.15. Thus, player 2 is better off.

I might make a another version of this puzzle where collusion is allowed, but let's wait for this one to be closed first.

Correct!!

Well done.

Presumably this game is a sort of "tortoise & hare" type of lesson teaching traders that opportunity can be found in unexpected places. Player #1 gets first dibs, receives insider information and only plays at an advantage. And yet it's blind Player #2--armed only with a second hand discount--who will be enjoying his Florida retirement long before Player #1 ever starts working on his suntan.

Gordon makes an interesting point about Player #1 playing hardball. I suppose a mathematician might look at it along the lines of the Nash equilibrium. Other ways to phrase it would be: kickback, shakedown, extortion, racketeering, greasing the wheels, corruption and cost of doing business.

-----------------------------------------

Because you lose half of your money but your wife is still there.
Have you tried 22 tonight? I said 22.
Gialmere
Joined: Nov 26, 2018
• Posts: 2472
November 24th, 2022 at 11:06:15 AM permalink
Happy (U.S.) Thanksgiving! Here's a few easy puzzles to let you escape your in-laws for a few moments...

John must play three singles matches against his parents. If he wins two matches in a row he gets money for a new computer.

His mother says he can play either Mother-Father-Mother, Father-Mother-Father or Father-Father-Mother.

His Father is a better player than his mother.

Which sequence should he play or are all three the same?

There are three pucks on the ice. A player shoots one of them between the other two. He repeats the shots never shooting the same puck twice in a row.

What is the minimum number of shots it will take to return the pucks back to their starting positions on the ice, or is that impossible?
Have you tried 22 tonight? I said 22.
Wizard
Joined: Oct 14, 2009
• Posts: 24887
Thanks for this post from:
November 24th, 2022 at 11:58:43 AM permalink
Quote: Gialmere

Happy (U.S.) Thanksgiving! Here's a few easy puzzles to let you escape your in-laws for a few moments...

John must play three singles matches against his parents. If he wins two matches in a row he gets money for a new computer.

His mother says he can play either Mother-Father-Mother, Father-Mother-Father or Father-Father-Mother.

His Father is a better player than his mother.

Which sequence should he play or are all three the same?

He should play FMF.

If the sequence is MFM, his chance of winning is 2mf-fm^2
If the sequence is MFM, his chance of winning is 2mf-mf^2

2mf-fm^2 > 2mf-mf^2
-fm^2 > -mf^2
fm^2 < mf^2
m < f (which we're given)

In more simple English, he must win game #2. Assuming he does, he gets two cracks at the other parent. Better to have to shots at the better player.

By the way, this one was asked in this month's Mensa newsletter. Are you a member, G?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
charliepatrick
Joined: Jun 17, 2011
• Posts: 2712
Thanks for this post from:
November 24th, 2022 at 1:42:14 PM permalink
I think it can be done with six shots. Consider a six sided star which comprises of two equalateral triangles. A is in the centre and B starts at (say) the SE point with C on the E point. B goes to the NE, C goes to the NW, B goes to the W. B is now opposite where C started, and C is oposite where B started. So continue the next three shots to get back to the original.

The solution relies on the fact that if all three move then each of them would need to go to somewhere and then back (i.e. six). But it is easier if one of them stays still and so the other two have to go round. The minimum for that is 3-moves each as each of them has to form a triangle.
Gialmere
Joined: Nov 26, 2018
• Posts: 2472
November 24th, 2022 at 10:50:37 PM permalink
Quote: Wizard

Quote: Gialmere

Happy (U.S.) Thanksgiving! Here's a few easy puzzles to let you escape your in-laws for a few moments...

John must play three singles matches against his parents. If he wins two matches in a row he gets money for a new computer.

His mother says he can play either Mother-Father-Mother, Father-Mother-Father or Father-Father-Mother.

His Father is a better player than his mother.

Which sequence should he play or are all three the same?

He should play FMF.

If the sequence is MFM, his chance of winning is 2mf-fm^2
If the sequence is MFM, his chance of winning is 2mf-mf^2

2mf-fm^2 > 2mf-mf^2
-fm^2 > -mf^2
fm^2 < mf^2
m < f (which we're given)

In more simple English, he must win game #2. Assuming he does, he gets two cracks at the other parent. Better to have to shots at the better player.

By the way, this one was asked in this month's Mensa newsletter. Are you a member, G?

Quote: charliepatrick

I think it can be done with six shots. Consider a six sided star which comprises of two equalateral triangles. A is in the centre and B starts at (say) the SE point with C on the E point. B goes to the NE, C goes to the NW, B goes to the W. B is now opposite where C started, and C is oposite where B started. So continue the next three shots to get back to the original.

The solution relies on the fact that if all three move then each of them would need to go to somewhere and then back (i.e. six). But it is easier if one of them stays still and so the other two have to go round. The minimum for that is 3-moves each as each of them has to form a triangle.

Both Correct!!

Well done.

Yes. The key is that the second game is a must win. Since MMF is not an option, FMF is the best choice.

Yes. Although the puzzle specifies you cannot shoot the same puck twice in a row, it doesn't say you MUST shoot all three pucks. So...

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As for Mensa...

...even though the box said 2-4 years.
Have you tried 22 tonight? I said 22.
Gialmere
Joined: Nov 26, 2018
• Posts: 2472
November 25th, 2022 at 12:32:31 PM permalink
For the weekend, here is a classic geometry puzzle by Catriona Shearer called...

Spike in the Hive

Two of the regular hexagons are identical; the third has area 10.

What’s the area of the red triangle?
Have you tried 22 tonight? I said 22.
charliepatrick
Joined: Jun 17, 2011
• Posts: 2712
Thanks for this post from:
November 25th, 2022 at 1:59:49 PM permalink
Quote: Gialmere

...What’s the area of the red triangle?...

I haven't checked my arithmetic but it does sound like a nice answer! (I've added the coloured text etc. to explain some of my unstated thoughts.)
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5541
November 25th, 2022 at 2:12:23 PM permalink

Label the vertices as shown.
Let x be the length of the sides of the two hexagons.
Angles FMN and MEG are 120°, so EMN and MEN are 60°, which means MEN is an equilateral triangle
ME = EF - MF = x - 10, so EM and MN are also x - 10
LN = LM - MN = 10 - (x - 10) = 20 - x
LK = KN - KL = x - (20 - x) = 2x - 20
Angles POL and JKL are 120°, so LOK and LKO are 60°, which means LOK is an equliateral triangle
LO = LK, which means 10 = 2x - 20, which means x = 15

Now, assign coordinates to each point, starting with P being on (0,0)
F is (0, 10 sqrt(3)), so A is (-15/2, 10 sqrt(3) + 15 sqrt(3) / 2), or (-15/2, 35 sqrt(3) / 2), and B is at (0, 25 sqrt(3))
O is at (10, 0), OK = ON = 10, so K is at (20, 0), which means J is at (35, 0), and I is at (85/2, 15 sqrt(3)/2)

Using the "bounding box" method with I in the lower right corner, I get the area of the triangle to be:
(50 x 35 sqrt(3) / 2) - (1/2 x 15/2 x 15 sqrt(3) / 2) - (1/2 x 85/2 x 35 sqrt(3)/2) - (1/2 x 10 sqrt(3) x 50) = 450 sqrt(3)

I have a feeling that's either correct or nowhere near the answer...

That's the answer if the smaller hexagon has side length 10.

A hexagon with side length 10 has area 6 x (1/2 x 10 x 10 sqrt(3) / 2) = 150 sqrt(3), so to find the correct area of the triangle, multiply the area of the "side length 10" triangle by 10 / 150 sqrt(3); the answer = 450 sqrt(3) x 10 / 150 sqrt(3) = 30.

Last edited by: ThatDonGuy on Nov 25, 2022
Gialmere
Joined: Nov 26, 2018
• Posts: 2472
November 26th, 2022 at 9:39:33 AM permalink
Quote: charliepatrick

I haven't checked my arithmetic but it does sound like a nice answer! (I've added the coloured text etc. to explain some of my unstated thoughts.)

Correct!!

Excellent.

There’s a little equilateral triangle in the middle between the hexagons, and its equilateralness means a big hex side is equal to a little hex side plus a triangle side (H = h + t, from the upper big hex), but also a big hex side is two little hex sides minus a triangle side (H = 2h - t, from the lower big hex), and that means t = h/2 and H = 3/2 h. So the area of a big hex is 9/4 x 10.

Now, the red triangle area: It’s half base times height, so it’s twice the area of a triangle with the same base and half the height, or twice the area of the same triangle sheared parallel to its base:

And that last triangle clearly is 1/3 the area of the hexagon it’s inscribed in. So A = 2/3 x 9/4 x 10 = 15.

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To get to the other hive.

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Shearer is best known for the felt tip marker way she presents her puzzles. Another of her signature moves is neat and tidy answers. Here's another one of her classics. This one called...

Fly the Flags

Squares of the same color are the same size.